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One of my homework problems is to prove that $\sum_{i=1}^n (2i-1)^2 = \frac{n(2n+1)(2n-1)}{3}$

I already completed the basis step

$[2(1)-1]^2 = 1 $

$\frac{(1)[2(1)+1][2(1)-1]}{3} = 1$

Then I assumed that the proposition was true for n=k+1 which leaves me with $\frac{k(2k+1)(2k-1)}{3}$ + $[2(k+1)-1]^2$ = $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$

However, I am stuck here, as the term $[2(k+1)-1]^2$ produces a constant term and $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$ (when the numerator is multiplied out) has no constants. Any ideas, guys?

Edits: After reading some of the comments I found my original mistake. I should have written $\frac{(k+1)[2(k+1)+1][2(k+1)-1]}{3}$ instead of $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$.

This leaves me with $\frac{k(2k+1)(2k-1)}{3}$ + $[2(k+1)-1]^2$ = $\frac{(k+1)[2(k+1)+1][2(k+1)-1]}{3}$

However, I am still lost on what to do next. I'm not having any luck combining these fractions, anyone got good ideas?

Mathdude388
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3 Answers3

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What you must prove is that $$\frac{k(2k+1)(2k-1)}3+[2(k+1)-1]^2=\frac{\color{red}{(k+1)}[2(k+1)+1][2(k+1)-1]}3$$

ajotatxe
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  • Thanks for noticing! I'm still having some trouble though, I edited the question to reflect where I am at in solving the problem now. – Mathdude388 May 21 '19 at 14:27
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What you have to prove is $$\frac{k(2k+1)(2k-1)}{3}+(2k+1)^2=\frac{(k+1)(2k+3)(2k+1)}{3}$$ The left-hand side is $$\frac{k(2k+1)(2k-1)+3(2k+1)^2}{3}=\frac{(2k+1)(2k+3)(k+1)}{3}$$ $$(2k+1)(2k^2-k+6k+3)$$ and $$(2k+3)(k+1)=2k^2-k+6k+3$$

Dr. Sonnhard Graubner
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  • Thanks for noticing. I still have no idea how to reach the right hand side of this equation from the left side though, I've tried combining the fractions but to no avail. – Mathdude388 May 21 '19 at 14:42
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Let $S_n=\sum_{i=0}^n(2i-1)^2=\frac{n(2n+1)(2n-1)}{3}$. Then $S_1=\sum_{i=0}^1(2i-1)^2=2$. But on the other hand, $S_1=\frac{1\cdot(2+1)(2-1)}{3}=1$. Contradiction !

I guess there may be some error in your question.

Zuo Ye
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